SOLVING LOGARITHMS - MADE EASY WITH SOLVED EXAMPLES, EXERCISES, LINKS TO FURTHER STUDY

Please study

Logarithm Formulas before Solving Logarithms

if you have not already done so.

Having the knowledge of the Formulae
is a prerequisite here.

Example 1 of Solving Logarithms :

Express the following as single logarithms to any base.
(i) (1⁄2) log (a + b) - (1⁄2) log (a - b)
(ii) log (a + b) + log (a² - ab + b²) - log (x - y) - log (x² + xy + y²)
(iii) (3⁄2) log x - (1⁄3) log y + (2⁄3) log z - (1⁄5) log a

Solution to Example 1 of Solving Logarithms :

Solution to 1(i) of Solving Logarithms :

Let P = (1⁄2) log (a + b) - (1⁄2) log (a - b)
We know in log of a power (see Formula 7) , the exponent multipliesthe log. Considering the reverse, If a quantity multiplies thelog, it becomes exponent.
∴ P = log {(a + b)^(1⁄2)} - log {(a - b)^(1⁄2)} We know, log of a quotient can be written as the difference of the logs of the numerator and denominator of the quotient (See Formula 6)
Considering the reverse, the difference of the logs of different quantitiescan be written as log of a quotient (the quantity whose log has negative sign, in the denominator) of the quantities.
= log {(a + b)^(1⁄2)}⁄{(a - b)^(1⁄2)}
= log {(a + b)⁄(a - b)}^(1⁄2). Ans.

Solution to 1(ii) of Solving Logarithms :

Let P = log (a + b) + log (a² - ab + b²) - log (x - y) - log (x² + xy + y²)
= log (a + b) + log (a² - ab + b²) - {log (x - y) + log (x² + xy + y²)}
We know, log of a product can be written as the sum of the logs of the factors of the product (See Formula 5)
Considering the reverse, the sum of the logs of different quantitiescan be written as log of product of the quantities.
∴ P = log {(a + b)(a² - ab + b²)} - log {(x - y)(x² + xy + y²)}
We know from Polynomials, (a + b)(a² - ab + b²) = a³ + b³;
and (x - y)(x² + xy + y²) = x³ - y³
∴ P = log (a³ + b³) - log (x³ - y³)
We know, log of a quotient can be written as the difference of the logs of the numerator and denominator of the quotient (See Formula 6)
Considering the reverse, the difference of the logs of different quantitiescan be written as log of a quotient (the quantity whose log has negative sign, in the denominator) of the quantities.
∴ P= log {(a³ + b³)⁄(x³ - y³)} Ans.

Solution to 1(iii) of Solving Logarithms :

Let P = (3⁄2) log x - (1⁄3) log y + (2⁄3) log z - (1⁄5) log a
We know in log of a power (see Formula 7) , the exponent multipliesthe log. Considering the reverse, If a quantity multiplies thelog, it becomes exponent.
∴ P = log x^(3⁄2) + log z^(2⁄3) - {log y^(1⁄3) + log a^(1⁄5)}
We know, log of a product can be written as the sum of the logs of the factors of the product (See Formula 5)
Considering the reverse, the sum of the logs of different quantitiescan be written as log of product of the quantities.
∴ P = log {x^(3⁄2) x z^(2⁄3)} - log {y^(1⁄3) x a^(1⁄5)}
We know, log of a quotient can be written as the difference of the logs of the numerator and denominator of the quotient (See Formula 6)
Considering the reverse, the difference of the logs of different quantitiescan be written as log of a quotient (the quantity whose log has negative sign, in the denominator) of the quantities.
∴ P = log [{x^(3⁄2) x z^(2⁄3)}⁄{y^(1⁄3) x a^(1⁄5)}] Ans.

Example 2 of Solving Logarithms :

Solve the equations
(i) 3^{(2x + 1)}.4^{(4x - 1)} = 36
(ii) 4^{(log₉ 3)} + 9^{(log₂ 4)} = 10^{(log_x 83)}

Solution to Example 2 of Solving Logarithms :

Solution to 2(i) of Solving Logarithms :

3^{(2x + 1)}.4^{(4x - 1)} = 36
Taking logarithms on both sides, we get
log {3^{(2x + 1)}.4^{(4x - 1)}} = log 36
⇒ log {3^{(2x + 1)}} + log {4^{(4x - 1)}} = log 36
We know in log of a power (see Formula 7) , the exponent multipliesthe log.
∴ (2x + 1) log 3 + (4x - 1) log 4 = log 36
⇒ 2x log 3 + 4x log 4 + log 3 - log 4 = log 36
⇒ 2x(log 3 + 2 log 4) = log 36 - log 3 + log 4
We know in log of a power (see Formula 7) , the exponent multipliesthe log. Considering the reverse, If a quantity multiplies thelog, it becomes exponent.
∴ L.H.S. = 2x(log 3 + log 4²)
We know, log of a product can be written as the sum of the logs of the factors of the product (See Formula 5)
Considering the reverse, the sum of the logs of different quantitiescan be written as log of product of the quantities.
We know, log of a quotient can be written as the difference of the logs of the numerator and denominator of the quotient (See Formula 6)
Considering the reverse, the difference of the logs of different quantitiescan be written as log of a quotient (the quantity whose log has negative sign, in the denominator) of the quantities.
Applying these to L.H.S. and R.H.S., we get
2x{log (3 x 4²)} = log (36 x 4⁄3) = log 48
⇒ 2x{log 48} = log 48 ⇒ 2x = 1 ⇒ x = 1⁄2. Ans.

Alternative Solution: This problem can be solved from Exponents knowledge without using logarithms as given below.
3^{(2x + 1)}.4^{(4x - 1)} = 36 ⇒ 3^(2x).3¹.4^(4x).4^(-1) = 36
⇒ (3)^2x.(4²)^2x.(3⁄4) = 36
⇒ 3^2x.(16)^2x = 36 x (4⁄3) = 48 ⇒ (3 x 16)^2x = 48
⇒ (48)^2x = 48 ⇒ 2x = 1 ⇒ x = 1⁄2. Ans.

Solution to 2(ii) of Solving Logarithms :

The given equation is 4^{(log₉ 3)} + 9^{(log₂ 4)} = 10^{(log_x 83)}
We have log₉ 3 = log_3² 3¹
We know, in log of a power to the base of another power (see Formula 9),the quotient of the exponents (exponent of the base in the denominator) multiplies the log.
∴ log₉ 3 = (1⁄2) log₃ 3
We know log of a quantity to the same base is 1. (See Formula 4)
∴ log₉ 3 = (1⁄2) (1) = 1⁄2;
log₂ 4 = log₂ 2²
We know in log of a power (see Formula 7) , the exponent multipliesthe log.
∴ log₂ 4 = 2 log₂ 2
We know log of a quantity to the same base is 1. (See Formula 4)
∴ log₂ 4 = 2 (1) = 2
∴ the given equation becomes 4^(1⁄2) + 9² = 10^{(log_x 83)}
⇒ 2 + 81 = 10^{(log_x 83)}
⇒ 83 = 10^{(log_x 83)} .......(i)
By converting the exponential form of (i) to logarithmic form (See Formula 1), we get
log_x 83 = log₁₀ 83
When two logarithms of same quantities are equal, their bases have to be equal.
∴ x = 10. Ans.
You can also apply Formula 2 to equation (i), to conclude x = 10.Ans.

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Exercise : Solving Logarithms

1. Solve 4^x + 2^{(2x - 1)} = 3^{(x + 1⁄2)} + 3^{(x - 1⁄2)}
   (i) by using logarithms (ii) without using logarithms.
   
2. If (log_y x.log_z x - log_x x) + (log_z y.log_x y - log_y y) + (log_x z.log_y z - log_z z) = 0,
   prove that xyz = 1.

For Answer to problem1 of the above exercise on
Solving Logarithms see at the bottom of the page.

Answers to Exercise : Solving Logarithms

(1) 3⁄2